The correct option is B 2√3 m/s2, 25.85m
Given,
velocity of particle for the first 5 sec, v=(4+4√t) m/s
As we know that acceleration of a particle is time derivative of its instantaneous velocity.a=dvdt ⇒a=d(4+4√t)dt
⇒a=2√t
And at time t=3.0 sec, acceleration is
a=2√3 m/s2
To find displacement: We know that velocity is the time derivative of displacement, we have to integrate as follows
v=dxdt=4+4√t
⇒dx=(4+4√t)dt
Integrating the above expression from time t=0 to t=3.0 sec, we get the displacement, which varies from xi=0 to xf=x, for these time instants.
∫xf=xxi=0dx=∫t=3t=0(4+4√t)dt
⇒[x−0]=[4t+83t3/2]30
⇒x=[12+13.85]
∴x≈25.85 m
Hence, option (b) is the correct answer.